Em tham khảo nha:

\(\begin{array}{l}
10)\\
a)\\
BaC{l_2} + N{a_2}S{O_4} \to BaS{O_4} + 2NaCl\\
{V_{{\rm{dd}}BaC{l_2}}} = \frac{{120}}{{1,2}} = 100ml = 0,1\,l\\
{n_{BaC{l_2}}} = 0,1 \times 1 = 0,1\,mol\\
{n_{N{a_2}S{O_4}}} = \frac{{200 \times 14,2\% }}{{142}} = 0,2\,mol\\
\dfrac{{0,2}}{1} > \dfrac{{0,1}}{1} \Rightarrow N{a_2}S{O_4}\text{ dư}\\
{n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,1\,mol\\
{n_{NaCl}} = 2{n_{BaC{l_2}}} = 0,2\,mol\\
{n_{N{a_2}S{O_4}}} \text{ dư}= 0,2 - 0,1 = 0,1\,mol\\
{m_{{\rm{dd}}spu}} = 120 + 200 - 0,1 \times 233 = 296,7g\\
{C_\% }N{a_2}S{O_4} \text{ dư}= \dfrac{{0,1 \times 142}}{{296,7}} \times 100\%  = 4,79\% \\
{C_\% }NaCl = \dfrac{{0,2 \times 58,5}}{{296,7}} \times 100\%  = 3,94\% \\
11)\\
{m_{{O_2}}} = m = 6,4g\\
{n_{{O_2}}} = \dfrac{{6,4}}{{32}} = 0,2\,mol\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
2KN{O_3} \xrightarrow{t^0} 2KN{O_2} + {O_2}\\
hh:KCl{O_3}(a\,mol),KN{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
1,5a + 0,5b = 0,2\\
122,5a + 101b = 22,35
\end{array} \right.\\
 \Rightarrow a = b = 0,1\\
\% {m_{KCl{O_3}}} = \dfrac{{0,1 \times 122,5}}{{22,35}} \times 100\%  = 54,81\% \\
\% {m_{KN{O_3}}} = 100 - 54,81 = 45,19\% 
\end{array}\)